Famous brain teaser

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It will be 50/50

It is either in one or the other.

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SFK

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I know that wiki page well, and for me the key sentence in that wiki page (as the Monty Hall problem has always hurt my brain) is:

Many readers of vos Savant's column refused to believe switching is beneficial and rejected her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them calling vos Savant wrong.[4] Even when given explanations, simulations, and formal mathematical proofs, many people still did not accept that switching is the best strategy.[5] Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant's predicted result.
 
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Its because the host knows where the car is and cannot choose it when he opens the alternative door. Therefore the odds of switching increase the chance of success, not because it somehow magically changes where the car is.

If there was no influence (i.e. the game restarted if the hose revealed the car by mistake) then there would be two independent choices. 1:3 1:2. with stick or change both having a 50/50 choice.
 
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Whilst the mathematical part is true, it takes no account of bluffing by the host.

Is he trying to make you change your mind to move you away from the car?
 
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He can't - he can only remove one dud door and must always offer stick or change.
 
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How can he do that.
- He must choose one dud door in round 1 and he must choose the non car door before offering the stick. He can therefore only choose 1 door unless the contestant has already made the correct choice.
 
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If there was no influence (i.e. the game restarted if the hose revealed the car by mistake) then there would be two independent choices. 1:3 1:2. with stick or change both having a 50/50 choice.

I've been trying to get my head around that point.

Say, the game was played, but the host doesn't know where the car is. The contestant chooses a door, and then the host opens a door which (by chance) reveals a booby prize. What would the odds then be on sticking or changing to win the car?
 
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I've been trying to get my head around that point.

Say, the game was played, but the host doesn't know where the car is. The contestant chooses a door, and then the host opens a door which (by chance) reveals a booby prize. What would the odds then be on sticking or changing to win the car?
It doesn't matter in that case. You've accidentally arrived at the exact same situation. You still have a 2:1 better chance switching at that stage.
 
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It doesn't matter in that case. You've accidentally arrived at the exact same situation. You still have a 2:1 better chance switching.

Thanks. I thought so (honestly!). But didn't want to commit.
 
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If nobody knew nuffink and didn't show you anything then you'd have a 1 in 3 chance.

It's 1) because a door is opened for you, and 2) because you know the host's job which tells you which he will pick in the given circumstance, that you can do better than 1 in 3.
Then it's not hard to add up the probabilities.
If there were 9 doors it wouldn't help you much; it's because there are only 3 that the activities make such a significant improvement in your odds.

It doesn't matter how much the odds improve, you just take it!
That's why it's almost intuitive - it takes imperceptible thought because you don't have to work anything out.
 
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I think it's still 50.50 on the remaining doors. All opening a don't win door does is reduce the odds to 50.50.

All opening the don't win does is shows that the car is behind either one of the two remaining doors is a win. The 3 doors were initially set up randomly. That nocks a hole in her argument.
 
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